blogfast25
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Determination of the alkali metals in silicious minerals.
Looking for ways to analyse some Lepidolites sample I have, I came across many references to an old method developed by famous American scientist and
chemist John Lawrence Smith published in Am. Jour. Sci., [2] xv, 234 - 243, in 1853 (!). Now undoubtedly obsolete, at the time it replaced Berzelius'
method of acid digestion (H2SO4 + HF).
Here's one reference to it, from 1949:
http://pubs.acs.org/doi/abs/10.1021/ac60028a015
And here's one from 1926 (shows also an interesting way of separating Rb and Cs, using 3CsCl.2SbCl3 and Rb2SnCl6:
http://www.journalarchive.jst.go.jp/jnlpdf.php?cdjournal=bcs...
The Lawrence Smith method appears to have been based on fusing the sample with 'carbonate of lime' (CaCO3) and using CaCl2 as fluxing agent.
Has anybody here have a precise description of the Lawrence Smith method of extracting alkali metals from silicate minerals?
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not_important
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You want:
Determination of the common and rare alkalies in mineral analysis H.H.Willard
http://www.archive.org/details/advancedquantita031019mbp
and/or from Google books:
Laboratory manual for the course in Advanced Quantitative Analysis Willard
Thumbnail is a dry mixture of very pure CaCO3 and NH4Cl is used, with enough CaCO3 to result in at least one mole excess of CaO during heating. A
tall crucible with a cap is used, the cap and cooler upper section of the crucible preventing loss of the alkalies as volatile chlorides. If there is
much sulfate in the sample BaCl2 is substituted for some to all of the NH4Cl, retaining the sulfate as BaSO4 ppt during the water extraction of the
cooled fusion mixture.
CaCO3 and NH4Cl are used because it is easier to obtain or produce them with low alkali metal content, especially in the days before AR grade
reagents. CaCO3 is precipitated by ammonium carbonate from fairly dilute solution, and well washed with water containing some NH3-NH4Cl-ammonium
carbonate. NH4Cl is purified by repeated sublimation.
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bahamuth
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Not the same substrate but one place I worked we digested ultra purified quartz in 40% HF and had it on heating(70C) in Teflon EM flasks with screw
cap overnight IIRC, and then let it evaporate and redissolved in 69% HNO3, evaporated and redissolved again, and dilution before we run it on a
ICP-MS.
Nasty but effective, though we probably lost some of the transition metals like titanium and the like to volatile fluoride species.
Alkaline and earth alkaline metals was our main concern..
Still remember the corroded windows and the ceramic hotplates....
Any sufficiently advanced technology is indistinguishable from magic.
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blogfast25
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Thanks, not_important, just the thing! Page 266 gives the full fusing procedure.
When reading up on Lepidolite, it's impossible not to come across the story of Bunsen and Kirchhoff's discovery of Rb in that mineral:
From Wiki:
"In 1861 Robert Bunsen and Gustav Kirchhoff extracted 150 kg of lepidolite and yielded a few grams of rubidium salts for analysis, and therefore
discovered the new element rubidium."
The Rb content of Lepidolite varies from about 0.3 to 3.5 %, so a 150 kg contains somewhere 450 g and 5 kg of Rb. It makes one wonder what kind of
extraction procedure they used to obtain only 'a few grams of rubidium salts!
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watson.fawkes
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Quote: Originally posted by blogfast25  | | The Rb content of Lepidolite varies from about 0.3 to 3.5 %, so a 150 kg contains somewhere 450 g and 5 kg of Rb. It makes one wonder what kind of
extraction procedure they used to obtain only 'a few grams of rubidium salts! |
Well, if you don't have any
idea that it's there, it's easy to wash it away in some other step of the process. Rather, I'd say it's lucky for them that they were using a process
that left any residue at all.
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blogfast25
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Watson.fawkes:
That's a little facile, IMHO: they were specifically looking for new trace elements of Group 1. Using a mineral that contained Li, Na and K was a
logical choice as hunting ground for any missing alkali metal elements. If you're extracting for trace metals, you'd be careful to avoid wastage...
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watson.fawkes
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Then I take it back. I don't know the
story, and it sounded serendipitous. Still, though, it's quite a problem to perform any kind of separation with reliability when you don't know the
properties of what you're looking for or its compounds.
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blogfast25
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On the latter we can certainly agree. Of course the statement regarding 'a few grams' (which is parroted by just about any other source) may simply
not be very accurate either. Only a reading of Bunsen and Kirchhoff's initial announcement of the work leading to the discovery may shed light on the
matter...
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JohnWW
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Lepidolite is a potassium lithium aluminium silicate, a micaceous mineral, associated with granite, and an important ore of Li. To separate the small
amount of Rb (and Cs) that partly replaces the K, and from any Na present, would require preferably some sort of ion-exchange chromatography.
Fractional crystallization, making use of differential solubilities, could be used, but it would be a much slower process. The Li would be capable of
being chemically separated, e.g. precipitation using the near-insolubility of LiF, the sparse solubility of which in water decreases with temperature
to only 0.13 gm/100 gm water, or of Li3PO4 which is even less soluble.
[Edited on 27-3-10 by JohnWW]
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not_important
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Bunsen and Kirchhoff new they had new elements from the spectral lines, and during the isolation process would check the two phases of a fractionation
operation (chloroplatinates I think) for the intensity of the target lines in each phase; the one with the most intense lines would be carried on.
It's possible that the less concentrated portion was not recycled back a step or two for further fractionation, but simply set aside. This would lead
to considerable inefficiencies with corresponding reduction in final yields.
There might be a bit more information at
http://www.jstor.org/pss/3905741
http://dx.doi.org/10.1021%2Fed009p1413
http://dx.doi.org/10.1002%2Fandp.18611890702
Until at least the mids-1990s Rb and Cs were isolated by fractional crystallisation, the alums being used and sometimes precipitating the complex
salts with SbCl3, TeCl4, or mixed halides M2ICl. I don't recall every coming across the use of ion exchange methods in commercial operations.
Attached is a Russian report on production of Rb and Cs as byproducts from several mining operations.
Attachment: Rb_Rus.pdf (350kB)
This file has been downloaded 726 times
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blogfast25
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| Quote: Originally posted by not_important | Bunsen and Kirchhoff new they had new elements from the spectral lines, and during the isolation process would check the two phases of a fractionation
operation (chloroplatinates I think) for the intensity of the target lines in each phase; the one with the most intense lines would be carried on.
It's possible that the less concentrated portion was not recycled back a step or two for further fractionation, but simply set aside. This would lead
to considerable inefficiencies with corresponding reduction in final yields.
|
Yes, but that seems so wasteful. Surely they would have used cascaded fractional crystallisation, rather than waste so much of the precious elements
and having to process up to 150 kg (!) of ore (for an alleged 'few grams')?
Perhaps their extraction technique wasn't very complete? Acid digestion, but not followed up with a carbonate/alkali fusion on the residue?
My whole interest in all this is mainly to observe the Rb spectral lines (and maybe the Cs ones too) using my homemade DVD spectroscope. I'm building
a new model that will allow also wavelength measurement, quite similar to Kirchhoff's instrument (but with grating instead of prism).
One source (which unfortunately I can't find back right away) I came across recently mentions an American prospector using an interesting technique to
generate the spectra. For that purpose he ground the mineral sample finely and mixed it with finely ground Fluorite. He then wetted the mixture a bit
and put some on the end of a charred match. The 'loaded' match then went into an acetylene flame and produced the spectra, even of trace elements.
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blogfast25
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Having had a closer look at the separation of RbCl and KCl this seems easier to achieve than might appear at first glance, at least if you have a good
a priori idea of the K/Rb ratio.
For this example, the quantities of KCl and RbCl were estimated from the theoretical composition of Lepidolite with an Rb content of 3 w%: KCl = 4.2
g, RbCl = 0.93 g (they are the quantities that would occur in 10 g of dried alkali salts from Lepidolite using the Lawrence Smith method).
I'm assuming here that LiCl, NaCl and CsCl were previously separated from the K/Rb.
Firstly we could take advantage of RbCl's greater solubility: at 20 C, water Sol. RbCl = 91 g / 0.1 L, Sol. KCl = 34.4 g / 0.1 L
Step 1:
To dissolve the 4.2 + 0.93 g sample the amount of solute needed is 4.2 g * 0.1 L / 34.4 g= 0.0122 L (12.2 mL)
To keep all the RbCl in solution we need the amount of solute = 0.93 g * 0.1 L / 91 g = 0.001 L (1 mL)
So by reducing the 12.2 mL total solution to about 1 mL most of the KCl crystallises out but all of the RbCl remains in the mother liquor. It contains
also 0.001 L * 34.4 g / 0.1 L = 0.344 g of KCl, so that after evaporating the decanted mother liquor the sample now contains KCl = 0.344 g, RbCl =
0.93 g
Step 2:
Enter the complex chlorides. Both hexachloro stannates and hexachloro platinates are suitable, the former being even more insoluble than the latter.
For the platinates, My Olde Holleman gave the following solubilities (at 10 C, water), expressed as the simple chlorides: Sol. K2PtCl6 = 0.28 g KCl /
0.1 L and Sol. Rb2PtCl6 = 0.064 g RbCl / 0.1 L
To dissolve 0.344 g KCl the amount of solute needed is 0.344 g * 0.1 L / 0.28 g = 0.123 L (123 mL) but this dissolves only 0.123 L * 0.064 g / 0.1 L =
0.0782 g, the rest of the RbCl remaining as crystals.
After filtering, the separation degree (RbCl yield) is thus 91.5 %.
Note that had the KCl/RbCl mother solution during step 1 for instance been reduced to 2 mL instead of 1 mL, the yield would have been 83.1 %.
Using the same procedure with hexachloro stannates calculates a yield of 95 %, presumably on account of the RbSnCl6 being even more insoluble and thus
less being lost to the mother liquor.
Zirconium and Hafnium were once also first separated (by de Hevesy and Coster) by means of fractional chrystallisation of the ammonium hexafluoro
zirconate and hafnate [(NH4)2ZrF6 and (NH4)2HfF6] but according to Holleman the Hafnium 'concentrates in the mother liquor', presumably that means
(NH4)2HfF6 is the more soluble of the two, as well as being the minority constituent.
Anybody have solubility data for (NH4)2ZrF6 and (NH4)2HfF6?
[Edited on 31-3-2010 by blogfast25]
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not_important
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I was sure I'd posted the solubility data for those or the potassium salts here before, but I can't find such :-(
I'll keep looking, the original source is here:
http://www3.interscience.wiley.com/journal/109796875/abstrac...
possibly of interest:
Inorganic Synthesis volume 3 includes isolating Cs compounds
http://rushim.ru/books/polytom/inorganic-synthesis/inorganic...
while vol. 4 has separating Hf and Zr by fractional ppt of the phosphates using a technique that produces them in a compact filtratable form instead
of gels
http://rushim.ru/books/polytom/inorganic-synthesis/inorganic...
A currently used process is based on the differing solubilities of the thiocyanates in MIBK.
[Edited on 1-4-2010 by not_important]
Attachment: Separation of Hafnium from Zirconium and Their Determination.pdf (183kB)
This file has been downloaded 499 times
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blogfast25
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not_important:
The values of solubility of the complex ammonium fluorides can almost certainly be found here:
van Arkel, A. E.; de Boer, J. H. (1924). "Die Trennung von Zirkonium und Hafnium durch Kristallisation ihrer Ammoniumdoppelfluoride (The
separation of zirconium and hafnium by crystallization of the double ammonium fluorides)" (in German). Zeitschrift für anorganische und allgemeine
Chemie 141: 284.
But this is a pay-for-view link:
http://www3.interscience.wiley.com/journal/109796875/abstrac...
What intrigues me about the ammonium double fluorides method is for that to work the Hf salt would have to be much, much more soluble than the Zr
salt, since as the former is the minority constituent... It would also work the other way around but Holleman seems to indicate the Hf concentrates in
the mother liquor.
Now I'll scrutinise your links, they sound very interesting....
[Edited on 2-4-2010 by blogfast25]
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